[EM] A more elegant three-candidate fpA-fpC phrasing, inspired by Heaviside formulation

[Kristofer Munsterhjelm]

For three candidates: Let A be the current candidate, then over all
other candidates B who A beats, add 2 fpA + fpB to A's score. Highest
score wins.

This works because in an A>B>C>A three-cycle, fpA - fpC = |V| + fpA -
fpC where |V| is the number of voters, because we're just adding a
constant. And |V| = fpA + fpB + fpC, so fpA - fpC = fpA + fpB + fpC +
fpA - fpC. The fpCs cancel and we get 2 fpA + fpB.

Now suppose that A is the Condorcet winner. Then A's score is 4 fpA +
fpB + fpC, i.e. 3 fpA + |V|. This exceeds the score any of the other
candidates can obtain, and so this version of fpA-fpC still passes
Condorcet.

In the case of a two-way tie, if we use the symmetric step function,
H(0) = 0.5, then we should have A's score be fpA + fpB/2 in such a case.
This would in effect break a two-way tie in favor of the Plurality
winner of the two.

f(A, e) = sum over B != A: H(A>B) * (2 fpA + fpB).

Unlike IRV, here the H term is a pairwise one. This makes it hard to
construct a recursive version. Now we could of course just generalize
the "sum over all B that A beats" concept to more than three candidates,
but I don't think the resulting method would pass DMTBR, and I also
think it would fail clone dependence (vote splitting with a cycle among
the clones).

-km